Answer
tangent $y=-\frac{5}{4}x+6$
normal $y=\frac{4}{5}x-\frac{11}{5}$
Work Step by Step
Step 1. Given $x+\sqrt {xy}=6$, we have $1+\frac{\sqrt y}{2\sqrt x}+\frac{\sqrt x}{2\sqrt y}y'=0$ or
$y'=-\frac{1+\frac{\sqrt y}{2\sqrt x}}{\frac{\sqrt x}{2\sqrt y}}=-\frac{2\sqrt {xy}+y}{x}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(4,1)$, the slope of the tangent line is $m=-\frac{2\sqrt {4}+1}{4}=-\frac{5}{4}$ and the equation is $y-1=-\frac{5}{4}(x-4)$ or $y=-\frac{5}{4}x+6$
Step 3. The slope of the normal line is $n=-1/m=\frac{4}{5}$ and the equation is $y-1=\frac{4}{5}(x-4)$ or $y=\frac{4}{5}x-\frac{11}{5}$
